A moving coil galvanometer has 150 equal divi | vedclass

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(B) The voltage sensitivity $(V_s)$ is related to current sensitivity $(I_s)$ and galvanometer resistance $(G)$ by the formula: $V_s = \frac{I_s}{G}$.

Vedclass · Accepted Answer

$9995$

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(B) The voltage sensitivity $(V_s)$ is related to current sensitivity $(I_s)$ and galvanometer resistance $(G)$ by the formula: $V_s = \frac{I_s}{G}$.Given $I_s = 10 \, 	ext{div/mA}$ and $V_s = 2 \, 	ext{div/mV}$,we find $G = \frac{I_s}{V_s} = \frac{10}{2} = 5 \, \Omega$.The full-scale deflection current $(I_g)$ is the total divisions divided by current sensitivity: $I_g = \frac{150 \, 	ext{div}}{10 \, 	ext{div/mA}} = 15 \, 	ext{mA} = 15 	imes 10^{-3} \, 	ext{A}$.To make each division read $1 \, V$,the total voltage $(V)$ to be measured for full-scale deflection is $150 \, 	ext{divisions} 	imes 1 \, 	ext{V/division} = 150 \, 	ext{V}$.To convert the galvanometer into a voltmeter of range $V$,a series resistance $R$ is required,given by $R = \frac{V}{I_g} - G$.Substituting the values: $R = \frac{150}{15 	imes 10^{-3}} - 5 = 10000 - 5 = 9995 \, \Omega$.

$A$ moving coil galvanometer has $150$ equal divisions. Its current sensitivity is $10$ divisions per milliampere and voltage sensitivity is $2$ divisions per millivolt. In order that each division reads $1 \, V$,the resistance in $\Omega$ needed to be connected in series with the coil will be

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